OTROS APORTES: SHAININ Y CROSBY

In the early days of control system design, controllers were usually implemented in analog form. Due to its simplicity, the phase lead-lag compensator was a popular form of controller since it can be easily implemented using a passive RC (resistor and capacitor) network or an RC network with an operational amplifier.

Basically, there are three commonly used compensators, namely, the phase lead com-pensator, phase lag comcom-pensator, and phase lead-lag compensator. Note that the compen-sator, or the controller, Gc(s), is usually applied in cascade (series) connection to the plant model G(s).

Phase lead compensator

The equivalent RC network to realize a phase lead compensator is shown in Figure 5.1(a).

We denote the impedances by Z₁= R1/(1+ R1Cs)and Z₂= R2. The transfer function of the phase lead network can be written as

Gc(s)= U_o(s)

U_i(s) = Z₂ Z₁+ Z2 = 1

α

1+ αT s

1+ T s , (5.1)

where

T = R₁R₂ R₁+ R2

C, α= R₁+ R2

R₂ . (5.2)

Obviously, α > 1. In general, the phase lead compensator can be written as G_c(s)= Kc

1+ αT s

1+ T s . (5.3)

The pole-zero location of the compensator is sketched in Figure 5.1(b). Since α > 1, the pole is always located on the left-hand side of the zero. For some different α’s, the Bode and Nyquist diagrams of the lead compensator with T = 1 are shown in Figures 5.2(a) and (b), respectively, through the following MATLAB statements:

R1 C Z₁

Z₂

?

ui

?

uo R₂

(a) lead network

-6^Im

− 1 Re

−1 αT T

(b) pole-zero positions

Figure 5.1. Lead compensator.

>> f1=figure; f2=figure; T=1;

for alpha0=1.5:0.5:5

G1=tf([alpha0*T,1]/alpha0,[T,1]);

figure(f1), nyquist(G1);hold on;figure(f2), bode(G1); hold on end

It can be observed that when α is large, the gain compensation is small but the phase compensation is large.

Example 5.1. Consider a plant model given by

G(s)= 100

s(0.04s+ 1).

The behavior of a feedback system with a lead compensator for the above plant model is illustrated in the frequency domain through this example.

The gain and phase margins of the system with the plant alone in the loop are obtained using the following MATLAB statements:

>> G=tf(100,[0.04,1,0]); [Gm,Pm,Wcg,Wcp]=margin(G), bode(G) It is found that the phase margin is 28.0243^◦at a frequency of 46.9701 rad/sec, with an infinite gain margin. The Bode diagram of the open-loop model is shown in Figure 5.3(a), where the phase margin is marked.

The phase margin can be increased by introducing a phase lead compensator given by G_c(s)= (0.0262s + 1)/(0.0106s + 1). The Bode diagram of the compensator is shown in Figure 5.3(b). In this case, the gain and phase margins of the compensated system can be obtained using the following MATLAB statements:

>> Gc1=tf([0.0262 1],[0.0106,1]); bode(Gc1) [Gm,Pm,Wcg,Wcp]=margin(G*Gc1)

It is found that the phase margin of the compensated system is 47.6^◦at a frequency of 60.3 rad/sec, again with an infinite gain margin. The magnitude and phase crossover frequencies

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Figure 5.2. Frequency domain representation of a lead compensator.

−60 Phase Margin (deg): 28 Delay Margin (sec): 0.0104 At frequency (rad/sec): 47 Closed Loop Stable? Yes

Phase (deg)

(a) Plant Bode diagrams

0

(b) Compensator Bode diagrams

Figure 5.3. Bode diagrams of the plant and the lead compensator.

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Phase Margin (deg): 47.6 Delay Margin (sec): 0.0138 At frequency (rad/sec): 60.3 Closed Loop Stable? Yes

Phase (deg)

(a) Bode diagrams comparison

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45

(b) step response comparison

Figure 5.4. Comparison of system responses.

in the compensated system, as expected, are both increased. The open-loop Bode diagrams of the compensated system and the original system are compared in Figure 5.4(a) using the following MATLAB statements:

>> G_o=Gc1*G; bode(G,G_o); figure

G_c1=feedback(G,1); G_c2=feedback(G_o,1); step(G_c1,G_c2)

The closed-loop step responses of the systems before and after phase lead compen-sation are compared in Figure 5.4(b). The step response of the compensated system is significantly improved, since the overshoot is reduced due to the increased phase margin, and the speed of response is also increased, due to the increased crossover frequency.

Phase lag compensator

The equivalent RC network for a phase lag compensator is shown in Figure 5.5(a), with the pole-zero positions sketched in Figure 5.5(b). Let Z1 = R1and Z2 = R2+ 1/(Cs). The

R1

Figure 5.5. Lag compensator.

0

Figure 5.6. Frequency responses of lag compensators.

transfer function of the phase lag network can be written as G_c(s)=U_o(s) compensator can be written as

Gc(s)= Kc

1+ αT s

1+ T s . (5.5)

The Bode diagrams and the Nyquist plots for K_c = 1 and T = 1 are shown in Figures 5.6(a) and (b), respectively, for different values of α. These diagrams are obtained using the following MATLAB statements:

>> f1=figure; f2=figure; T=1;

for alpha0=0.9:-0.1:0.1

G1=tf([alpha0*T,1]/alpha0,[T,1]);

figure(f1), nyquist(G1), hold on;figure(f2), bode(G1), hold on end

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−50 0 50 100

Magnitude (dB)

Bode Diagram

Frequency (rad/sec)

10⁻² 10⁻¹ 10⁰ 10¹ 10² 10³

−180

−135

−90 System: G_o

Phase Margin (deg): 50.8 Delay Margin (sec): 0.0529 At frequency (rad/sec): 16.7 Closed Loop Stable? Yes

Phase (deg)

← original system

compensated→

original system →

(a) Bode diagrams comparison

0 0.1 0.2 0.3 0.4 0.5

0 0.5 1 1.5

Step Response

Time (sec)

Amplitude

← uncompensated system

← with lag compensator

← with lead compensator

(b) step response comparison

Figure 5.7. Comparison of system responses.

Example 5.2. Consider again the plant model in Example 5.1. If a phase lag compensator G_c(s)= (0.5s +1)/ (2.5s +1) is now used, the gain and phase margins of the compensated system can be obtained using the following MATLAB statements:

>> Gc2=tf([0.5 1],[2.5,1]); G=tf(100,[0.04,1,0]); G_o=Gc2*G;

[Gm,Pm,Wcg,Wcp]=margin(G_o); bode(G_o,G)

figure;step(feedback(G,1),feedback(G_o,1),feedback(Gc1*G,1),0.5) The phase margin is 50.7572^◦at a frequency of 16.7339 rad/sec, with an infinite gain margin.

The Bode diagram of the compensated system can be obtained as shown in Figure 5.7(a).

The basic idea of a lag compensator is to decrease the crossover frequency so as to increase the phase margin of the system. However, since this technique reduces the open-loop bandwidth, it also reduces the response speed of the system. However, it does have the advantage, unlike lead compensation, that a solution can always normally be found.

The step responses of the phase lag compensated system, the original system, and the phase lead compensated system, are all shown in Figure 5.7(b). As with the lead compensator, the increased phase margin given by the lag compensator has reduced the overshoot in the step response.

Now, let us fix α = 0.2 and change T , i.e., the lag compensator is Gc(s) = (1 + 0.2T s)/(1+ T s). Let us see how T affects the performance of the compensated system.

Using the MATLAB statements

>> G=tf(100,[0.04,1,0]); f1=figure; f2=figure;

for T=[0.5,1,2.5,5,10,20]

Gc2=tf([0.2*T 1],[T,1]); G_o=G*Gc2; G_c=feedback(G_o,1);

figure(f1),bode(G_o), hold on;figure(f2),step(G_c,1), hold on end

the superimposed Bode diagrams and step response are shown in Figures 5.8(a) and (b), respectively. Among the phase lag compensators, it can be seen that the larger the value of T, the better the performance of the compensated system.

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(a) Bode diagrams comparison

0 0.2 0.4 0.6 0.8 1

(b) step response comparison

Figure 5.8. The effect of changing T .

R₁

Figure 5.9. Lead-lag compensator.

Phase lead-lag compensator

For a phase lead-lag compensator, its equivalent RC network is shown in Figure 5.9(a), and its pole-zero map is shown in Figure 5.9(b). Denote Z₁ = R1/(1+ R1C₁s) and Z2 = R2+ 1/(C2s). The transfer function of the phase lead-lag compensator can be written as property, while the second term has the phase lag characteristics.

As shown in Figure 5.10, with T₁ = 0.5, T2 = 0.005, and α = 3, β = 1/3, and T₂ = 0.5, T1 = 0.005, the Bode diagrams of the lead-lag compensator and the lag-lead compensators are obtained using the following MATLAB statements, respectively:

>> T1=0.5; T2=0.005; alpha=3; beta=1/3; s=zpk(’s’);

G1=(alpha*T1*s+1)*(beta*T2*s+1)/(T1*s+1)/(T2*s+1);

T2=0.5; T1=0.005; alpha=3; beta=1/3;

G2=(alpha*T1*s+1)*(beta*T2*s+1)/(T1*s+1)/(T2*s+1); bode(G1,G2)

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